A079360 Sequence of sums of alternating increasing powers of 2.

1, 5, 7, 15, 19, 35, 43, 75, 91, 155, 187, 315, 379, 635, 763, 1275, 1531, 2555, 3067, 5115, 6139, 10235, 12283, 20475, 24571, 40955, 49147, 81915, 98299, 163835, 196603, 327675, 393211, 655355, 786427, 1310715, 1572859, 2621435, 3145723

Offset: 0

Author

Cino Hilliard, Feb 15 2003

Keywords

• easy
• nonn

Comments

Found as a question on http://mail.python.org/mailman/listinfo/tutor poster: reavey.

Links

• G. C. Greubel, Table of n, a(n) for n = 0..5000
• Tutor User "reavey" ([e-mail]) and others, How to write an algorithm for sequence, Tutor -- Discussion for learning programming with Python, 2003.
• Index entries for linear recurrences with constant coefficients, signature (1,2,-2).

Crossrefs

Cf. A079361, A079362, A048488 (bisection).

Programs

GAP

a:=[1,5,7];; for n in [4..30] do a[n]:=a[n-1]+2*a[n-2]-2*a[n-3]; od; a; # G. C. Greubel, Aug 07 2019

Magma

I:=[1,5,7]; [n le 3 select I[n] else Self(n-1) +2*Self(n-2) -2*Self(n-3): n in [1..40]]; // G. C. Greubel, Aug 07 2019

Maple

seq(coeff(series((1+4*x)/((1-x)*(1-2*x^2)), x, n+1), x, n), n = 0..40); # G. C. Greubel, Aug 07 2019

Mathematica

LinearRecurrence[{1,2,-2}, {1,5,7}, 40] (* G. C. Greubel, Aug 07 2019 *)

PARI

seq(n) = { j=a=1; p=2; print1(1" "); while(j<=n, a = a + 2^p; print1(a" "); a = a+2^(p-1); print1(a" "); p+=1; j+=2; ) }

PARI

a(n)=if(n<0,0,(6-n%2)*2^ceil(n/2)-5)

Sage

@CachedFunction
def a(n):
    if (n==0): return 1
    elif (1<=n<=2): return nth_prime(n+2)
    else: return a(n-1) + 2*a(n-2) - 2*a(n-3)
[a(n) for n in (0..40)] # G. C. Greubel, Aug 07 2019

Formula

a(2n) = 6*2^n - 5, a(2n-1) = 5*(2^n - 1). - Benoit Cloitre, Feb 16 2003

From Colin Barker, Sep 19 2012: (Start)

a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3).

G.f.: (1+4*x)/((1-x)*(1-2*x^2)). (End)